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Jun 11, 2011 at 16:20 comment added Kevin Buzzard [s/subgroup/cyclic subgroup/ in the above]
Jun 11, 2011 at 16:19 comment added Kevin Buzzard So one way of doing a level $H$ structure (assuming $n$ invertible on the base) is to look at the isom scheme and then quotient out, and then observe that even if the isom scheme doesn't have points, the quotient still might. If you want to make this "explicit" you might start allowing base changes so you can see the points moving around under $H$.
Jun 11, 2011 at 16:17 comment added Kevin Buzzard I think you're a bit muddled. A full level $n$ structure is 1. I am not sure what you mean by 2 -- is the cover part of the data for example? -- but my guess is that there's been a misunderstanding here. The reason that one has to be a bit more careful when doing the general $H$ case is because you want to allow the situation that the curve have a level $H$ structure but have no full level $n$ structures at all -- e.g. if $H=\Gamma_0(n)$ then you want an $H$-structure to mean a subgroup of order $n$, but a curve can have such a subgroup without having a full level $n$ structure.
Jun 11, 2011 at 14:23 history asked unknown CC BY-SA 3.0